Question

我的数据框具有以下列结构（总计超过1000行）：

addressfull
POINT(3.124537653 32.179354012)||DEF_32||molengraaf 20, 1689 GL Utrecht, Netherlands||15||map
POINT(3.124537680 32.179354014)||DEF_32||winkellaan 67, 5788 BG Amsterdam, Netherlands||13||map
POINT(3.124537653 32.179354012)||DEF_32||vermeerstraat 18, 0932 DC Rotterdam, Netherlands||11||map
POINT(2.915206183 24.315583523)||DEF_32||--||13||map
POINT (2.900824999999923 34.3175721)||DEF_84||Zandhorstlaan 122, 0823 GT Ochtrup, Germany||17||map

structure(list(addressfull = structure(c(3L, 5L, 4L, 2L, 1L), .Label = c("POINT (2.900824999999923 34.3175721)||DEF_84||Zandhorstlaan 122, 0823 GT Ochtrup, Germany||17||map", 
"POINT(2.915206183 24.315583523)||DEF_32||--||13||map", "POINT(3.124537653 32.179354012)||DEF_32||molengraaf 20, 1689 GL Utrecht, Netherlands||15||map", 
"POINT(3.124537653 32.179354012)||DEF_32||vermeerstraat 18, 0932 DC Rotterdam, Netherlands||11||map", 
"POINT(3.124537680 32.179354014)||DEF_32||winkellaan 67, 5788 BG Amsterdam, Netherlands||13||map"
), class = "factor")), class = "data.frame", row.names = c(NA, 
-5L))

该列包含位置，街道，门牌号，邮政编码，城市和国家。我想在多个列中用R拆分addressfull列，例如：

street        house number      zip       city      country
molengraaf    20                1689 GL   Utrecht   Netherlands
winkellaan    67                5788 BG   Amsterdam Netherlands
vermeerstraat 18                0932 DC   Rotterdam Netherlands
na            na                na        na        na
Zandhorstlaan 122               0823 GT   Ochtrup   Germany

我已经阅读了tidyr和stringer文档。我可以看到用于从位置x分割（由“）”，“ |”和“，”分割的模式。但我找不到正确的代码将列拆分为多列。

有人可以帮助我吗？

Answer 1

对于基本的R方法，您可以使用sub对其进行暴力破解：

df$steet <- sub("^(\\S+)\\s+.*$", "\\1", df$adressfull)
df$`house number` <- sub("^\\S+\\s+(\\d+).*$", "\\1", df$adressfull)
df$zip <- sub("^\\S+\\s+\\d+,\\s*(\\d+\\s+[A-Z]+).*$", "\\1", df$adressfull)
df$city <- sub("^.*?(\\S+),\\s*\\S+$", "\\1", df$adressfull)
df$country <- sub("^.*,\\s*(\\S+)$", "\\1", df$adressfull)
df

                                   adressfull      steet house number     zip
1 molengraaf 20, 1689 GL Utrecht, Netherlands molengraaf           20 1689 GL
     city     country
1 Utrecht Netherlands

数据：

df <- data.frame(adressfull=c("molengraaf 20, 1689 GL Utrecht, Netherlands"),
                 stringsAsFactors=FALSE)

这假设我们已经隔离了地址文本。为此，请考虑：

text <- "POINT(3.124537653 32.179354012)||DEF_32||molengraaf 20, 1689 GL Utrecht, Netherlands||15||map"
addresfull <- unlist(strsplit(text, "\\|\\|"))[3]
addresfull

[1] "molengraaf 20, 1689 GL Utrecht, Netherlands"

Answer 2

这将是解决问题的一种整洁方法：

library(tidyverse)
df <- structure(list(addressfull = structure(c(3L, 5L, 4L, 2L, 1L), .Label = c("POINT (2.900824999999923 34.3175721)||DEF_84||Zandhorstlaan 122, 0823 GT Ochtrup, Germany||17||map", 
                                                                               "POINT(2.915206183 24.315583523)||DEF_32||--||13||map", "POINT(3.124537653 32.179354012)||DEF_32||molengraaf 20, 1689 GL Utrecht, Netherlands||15||map", 
                                                                               "POINT(3.124537653 32.179354012)||DEF_32||vermeerstraat 18, 0932 DC Rotterdam, Netherlands||11||map", 
                                                                               "POINT(3.124537680 32.179354014)||DEF_32||winkellaan 67, 5788 BG Amsterdam, Netherlands||13||map"
), class = "factor")), class = "data.frame", row.names = c(NA, 
                                                           -5L))



df %>% separate(addressfull, sep = "\\|\\|", into = c("Coords", "DEF", "ADDRESS"),extra = "drop") %>% 
  select(ADDRESS) %>% 
  separate(ADDRESS, sep = ",", into = c("street", "city", "country")) %>% 
  separate(street, sep = "(?= \\d)", into = c("street", "house_number")) %>% 
  separate(city, sep = "(?<=[A-Z][A-Z])", into = c("zip", "city"))
#> Warning: Expected 3 pieces. Missing pieces filled with `NA` in 1 rows [4].
#> Warning: Expected 2 pieces. Missing pieces filled with `NA` in 1 rows [4].
#>          street house_number      zip       city      country
#> 1    molengraaf           20  1689 GL    Utrecht  Netherlands
#> 2    winkellaan           67  5788 BG  Amsterdam  Netherlands
#> 3 vermeerstraat           18  0932 DC  Rotterdam  Netherlands
#> 4            --         <NA>     <NA>       <NA>         <NA>
#> 5 Zandhorstlaan          122  0823 GT    Ochtrup      Germany

由reprex软件包（v0.3.0）于2020-02-27创建

Answer 3

stringr解决方案是这样：

addresssplit <- data.frame(
  street = str_extract(addressfull$addressfull, "(?<=DEF_\\d{2}\\|\\|)\\w+\\b"),
  number = str_extract(addressfull$addressfull, "\\d{1,}(?=,)"),
  zip = str_extract(addressfull$addressfull, "(?<=\\s)\\d{4}\\s[A-Z]{2}"),
  city = str_extract(addressfull$addressfull, "(?<=\\d{4}\\s[A-Z]{2}\\s)\\w+"),
  country = str_extract(addressfull$addressfull, "(?<=[a-z]\\b,\\s)\\w+\\b")
)

结果：

addresssplit
         street number     zip      city     country
1    molengraaf     20 1689 GL   Utrecht Netherlands
2    winkellaan     67 5788 BG Amsterdam Netherlands
3 vermeerstraat     18 0932 DC Rotterdam Netherlands
4          <NA>   <NA>    <NA>      <NA>        <NA>
5 Zandhorstlaan    122 0823 GT   Ochtrup     Germany

数据：

addressfull <- structure(list(addressfull = structure(c(3L, 5L, 4L, 2L, 1L), .Label = c("POINT (2.900824999999923 34.3175721)||DEF_84||Zandhorstlaan 122, 0823 GT Ochtrup, Germany||17||map", 
                                                                                    "POINT(2.915206183 24.315583523)||DEF_32||--||13||map", "POINT(3.124537653 32.179354012)||DEF_32||molengraaf 20, 1689 GL Utrecht, Netherlands||15||map", 
                                                                                    "POINT(3.124537653 32.179354012)||DEF_32||vermeerstraat 18, 0932 DC Rotterdam, Netherlands||11||map", 
                                                                                    "POINT(3.124537680 32.179354014)||DEF_32||winkellaan 67, 5788 BG Amsterdam, Netherlands||13||map"
), class = "factor")), class = "data.frame", row.names = c(NA, 
                                                           -5L))

将完整地址列拆分为多列

3 个答案: